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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Indefinite Integration

NDA 2017 · Mathematics · Indefinite Integration

Let f(x) be an indefinite integral of ^2 x . Consider the following...

Let f(x)f(\mathrm{x}) be an indefinite integral of sin⁡2x\sin ^2 \mathrm{x}. Consider the following statements : Statement 1 : The function f(x)f(\mathrm{x}) satisfies f(x+π)=f(x)f(\mathrm{x}+\pi)=f(\mathrm{x}) for all real x. Statement 2:sin⁡2(x+π)=sin⁡2x2: \sin ^2(x+\pi)=\sin ^2 x for all real xx. Which one of the following is correct in respect of the above statements?

Options

  1. ABoth the statements are true and Statement 2 is the correct explanation of Statement 1
  2. BBoth the statements are true but Statement 2 is not the correct explanation of Statement 1
  3. CStatement 1 is true but Statement 2 is false
  4. DStatement 1 is false but Statement 2 is false

Correct answer

D. Statement 1 is false but Statement 2 is false

Step-by-step solution

f(x)=∫sin⁡2x⋅dx=∫1−cos⁡2x2⋅dx=12x−sin⁡2x2⋅12+c=12x−14sin⁡2x+c\begin{aligned} & f(x)=\int \sin ^2 x \cdot d x=\int \frac{1-\cos 2 x}{2} \cdot d x \\ & =\frac{1}{2} x-\frac{\sin 2 x}{2} \cdot \frac{1}{2}+c=\frac{1}{2} x-\frac{1}{4} \sin 2 x+c\end{aligned} (i) f(π+x)=12(π+x)−14sin⁡2(π+x)=12π+12x−12sin⁡(2π+2x)+c=x2−12sin⁡2x+c=f(x)\begin{aligned} & f(\pi+x)=\frac{1}{2}(\pi+x)-\frac{1}{4} \sin 2(\pi+x) \\ & =\frac{1}{2} \pi+\frac{1}{2} x-\frac{1}{2} \sin (2 \pi+2 x)+c \\ & =\frac{x}{2}-\frac{1}{2} \sin 2 x+c=f(x) \end{aligned} So, statement 1 is true. (ii) sin⁡2(π+x)=sin⁡2x(−sin⁡x)2=sin⁡2x⇒sin⁡2x=sin⁡2x\begin{aligned} & \sin ^2(\pi+x)=\sin ^2 x \\ & (-\sin x)^2=\sin ^2 x \\ & \Rightarrow \sin ^2 x=\sin ^2 x \end{aligned} So, statement 2 is true
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